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Homework Help: Order of variables in a Jacobian?

  1. Dec 12, 2012 #1

    1. The problem statement, all variables and given/known data

    I was hoping someone could please explain the order of variables in a Jacobian. I mean, once the dependent and independent variables have been identified, how should the Jacobian be formulated. For instance, supposing I have two implicit functions F(x,y,u,v) and G(x,y,u,v) where x and y are independent. I wish to find ∂u/∂y (x is fixed) and ∂v/∂y (x is fixed). How should the Jacobian be formulated, namely -J[(F,G)/(y,u)]/J[(F,G)/(v,u)] or -J[(F,G)/(u,y)]/J[(F,G)/(u,v)]? I happen to know the former formulation is the correct one, but why?! How could I have known and initially written it in that form?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 12, 2012 #2
    Consider two functions ##f## and ##g## of two variables y and x.
    I always form my Jacobian like
    \mathcal{J} = \begin{pmatrix}
    f_x & f_y\\
    g_x & g_y
    However, I believe in some books it has my rows as columns.
  4. Dec 12, 2012 #3


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    What's usually important is the absolute value of the determinant of the jacobian matrix. In which case none of these variations matter.
  5. Dec 12, 2012 #4

    Ray Vickson

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    You can always get it from first principles (and doing this once in your life is a useful exercise). If we fix y and let x change by Δx = h, then we have:
    [tex] F(x+h,y,u + \Delta u, v + \Delta v) = 0 = F(x,y,u,v) + F_x h + F_u \Delta u + F_v \Delta v \\
    G(x+h,y,u + \Delta u, v + \Delta v) = 0 = G(x,y,u,v) + G_x h + G_u \Delta u + G_v \Delta v[/tex]
    where all the partials are evaluated at the original point (x,y,u,v). Thus,
    [tex] \pmatrix{ \Delta u \\ \Delta v} =
    - \pmatrix{F_u & F_v\\G_u&G_v}^{-1}\pmatrix{F_x \\ G_x} h,[/tex]
    [tex] \pmatrix{\partial u/ \partial x \\ \partial v / \partial x} =
    - \pmatrix{F_u & F_v\\G_u&G_v}^{-1}\pmatrix{F_x \\ G_x}.[/tex]
    For a 2x2 matrix we get the inverse by swapping the diagonal elements, changing the sign of the off-diagonal elements and dividing by the determinant:
    [tex] \pmatrix{a & b \\ c & d }^{-1} = \frac{1}{ad-bc} \pmatrix{d & -b \\ -c & a},[/tex] so you can get explicit formulas for u_x and v_x.
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